A square-grid cycle enclosing a compact set

By [L1], choose a compact Jordan set $J$ such that $K\subseteq \mathrm{int}J\subseteq J\subseteq U$, and write $J$ as a finite union of closed cells from one square grid. Give every cell boundary its positive orientation. Each edge internal to $J$ then occurs twice with opposite orientations; cancel those pairs and let $\mathrm{\Gamma }$ be the finite chain of the remaining oriented frontier edges. At every grid vertex the incoming and outgoing coefficients balance, so $\mathrm{\Gamma }$ is a cycle. Its trace is the frontier of $J$, hence ${\mathrm{\Gamma }}^{\ast }\subseteq U\setminus K$.

Let $p\in \mathrm{int}J$ lie on no grid line. Summing the positively oriented boundaries of all cells gives the same integral and index as $\mathrm{\Gamma }$, because the two orientations of every internal edge cancel by [L3]. For one grid cell $Q$, the four-edge continuous argument of $\zeta -p$ makes one positive turn when $p\in \mathrm{int}Q$ and returns with zero net turn when $p\mathrm{\notin }Q$; hence [L2] gives $n({\mathrm{\partial }}^{+}Q,p)=1$ in the first case and $0$ in the second. Exactly one cell containing $p$ contributes $1$, so additivity in [L3] gives $n(\mathrm{\Gamma },p)=1$.

Fix $z\in K$. Choose a disc $D(z,r)\subseteq \mathrm{int}J$ and a point $p\in D(z,r)$ on no grid line. The disc misses ${\mathrm{\Gamma }}^{\ast }$ and is connected, so local constancy in [L4] and step 2.1 give $n(\mathrm{\Gamma },z)=n(\mathrm{\Gamma },p)=1$.